Question: Simplify the following expression: $y = \dfrac{9x^2+22x- 15}{9x - 5}$
First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(9)}{(-15)} &=& -135 \\ {a} + {b} &=& &=& {22} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-135$ and add them together. Remember, since $-135$ is negative, one of the factors must be negative. The factors that add up to ${22}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-5}$ and ${b}$ is ${27}$ $ \begin{eqnarray} {ab} &=& ({-5})({27}) &=& -135 \\ {a} + {b} &=& {-5} + {27} &=& 22 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({9}x^2 {-5}x) + ({27}x {-15}) $ Factor out the common factors: $ x(9x - 5) + 3(9x - 5)$ Now factor out $(9x - 5)$ $ (9x - 5)(x + 3)$ The original expression can therefore be written: $ \dfrac{(9x - 5)(x + 3)}{9x - 5}$ We are dividing by $9x - 5$ , so $9x - 5 \neq 0$ Therefore, $x \neq \frac{5}{9}$ This leaves us with $x + 3; x \neq \frac{5}{9}$.